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Thread: Challenging Integral

  1. #1
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    Challenging Integral

    solve $\displaystyle \int \frac{dx}{ sin^5{x} + cos^5{x} } $


    and


    $\displaystyle \int \frac{dx}{ sin^7{x} + cos^7{x} } $
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  2. #2
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    Quote Originally Posted by simplependulum View Post

    solve $\displaystyle \int \frac{dx}{ sin^5{x} + cos^5{x} } $ and $\displaystyle \int \frac{dx}{ sin^7{x} + cos^7{x} } $
    i'll solve the first one only and i'll leave the second integral to you or whoever likes trouble! first we need the following magical identity:

    $\displaystyle \frac{1}{\sin^5 x + \cos^5 x} = \frac{4}{5} \left[\frac{1}{\sin x + \cos x} + \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \right].$

    now finding $\displaystyle \int \frac{dx}{\sin x + \cos x}$ is straightforward. to find $\displaystyle I= \int \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \ dx$ substitute: $\displaystyle \sin x - \cos x = t.$ then $\displaystyle (\sin x + \cos x) \ dx = dt$ and thus

    $\displaystyle I=\int \frac{t^2-2}{t^4 - 4t^2 - 1} \ dt.$ finally let $\displaystyle a=2+\sqrt{5}$ and $\displaystyle b=2-\sqrt{5}$ to get: $\displaystyle I=\frac{1}{2} \left[\int \frac{dt}{t^2-a} + \int \frac{dt}{t^2 - b} \right].$ note that $\displaystyle b < 0.$ now everything is nice and easy and you're good to go!



    Remark: i think this is true that for any $\displaystyle n \in \mathbb{N}$ we have $\displaystyle \frac{1}{\sin^{2n+1}x + \cos^{2n+1} x} = \frac{2n}{2n+1}\left[\frac{1}{\sin x + \cos x} + (\sin x + \cos x)R((\sin x - \cos x)^2) \right],$ where $\displaystyle R(z)$ is some rational function of $\displaystyle z$.

    actually i think $\displaystyle R(z)=\frac{p(z)}{q(z)},$ where both $\displaystyle p(z)$ and $\displaystyle q(z)$ are monic polynomials with integer coefficients! for example, in your problem: $\displaystyle p(z)=z-2, \ q(z)=z^2-4z-1.$ i also think that we

    always have $\displaystyle \deg q(z) = 1+\deg p(z).$ these claims are either wrong or easy to prove! ... let's just leave it as a challenge for whoever has time and patience to try it.

    by the way, the identity $\displaystyle \frac{1}{\sin^3x + \cos^3x}=\frac{2}{3} \left[\frac{1}{\sin x + \cos x} + \frac{\sin x + \cos x}{(\sin x - \cos x)^2 + 1} \right]$ shows that my claims are not based on nothing!
    Last edited by NonCommAlg; Jun 22nd 2009 at 07:27 AM. Reason: nothing important!
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  3. #3
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    My solution is to substitute $\displaystyle t = \sin{x}+\cos{x} $
    so,
    $\displaystyle dt = (\cos{x} - \sin{x} )dx $

    $\displaystyle \sin{x} \cos{x} = \frac{t^2 - 1}{2} $

    $\displaystyle dx = \frac{dx}{ \sqrt{ 2 - t^2 } }$

    Although it looks very troublesome , it become quite well after separating it into fractions .

    However , i think your method is better than what i am using .
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